Cambridge University Press A Primer of Analytic Number Theory: From Pythagoras to Riemann Jeffrey Stopple Excerpt More information

Transcription

1 Chapter 1 Sums and Differences Imet a traveller from an antique land Who said: Two vast and trunkless legs of stone Stand in the desert Near them, on the sand, Half sunk, a shattered visage lies Percy Bysshe Shelley 11 Polygonal Numbers The Greek word gnomon means the pointer on a sundial, and also a carpenter s square or L-shaped bar The Pythagoreans, who invented the subject of polygonal numbers, also used the word to refer to consecutive odd integers: 1, 3, 5, 7, The Oxford English Dictionary s definition of gnomon offers the following quotation, from Thomas Stanley s History of Philosophy in 1687 (Stanley, 1978): Odd Numbers they called Gnomons, because being added to Squares, they keep the same Figures; so Gnomons do in Geometry In more mathematical terms, they observed that n is the sum of the first n consecutive odd integers: 1 = 1, =, = 3, = 4, Figure 11 shows a geometric proof of this fact; observe that each square is constructed by adding an odd number (the black dots) to the preceding square These are the gnomons the quotation refers to 1

2 1 Sums and Differences Figure 11 A geometric proof of the gnomon theorem But before we get to squares, we need to consider triangles The triangular numbers, t n, are the number of circles (or dots, or whatever) in a triangular array with n rows (see Figure 1) Since each row has one more than the row above it, we see that t n = n 1 + n A more compact way of writing this, without the ellipsis, is to use the Sigma notation, t n = k The Greek letter denotes a sum; the terms in the sum are indexed by integers between 1 and n, generically denoted k And the thing being summed is the integer k itself (as opposed to some more complicated function of k) Of course, we get the same number of circles (or dots) no matter how we arrange them In particular we can make right triangles This leads to a clever proof of a closed-form expression for t n, that is, one that does not require doing the sum Take two copies of the triangle for t n, one with circles and one with dots They fit together to form a rectangle, as in Figure 13 Observe that the rectangle for two copies of t n in Figure 13 has n + 1 rows and n columns, so t n = n(n + 1), or n(n + 1) n = t n = (11) This is such a nice fact that, we will prove it two more times The next proof is more algebraic and has a story The story is that Gauss, as a young student, was set the task of adding together the first hundred integers by his teacher, with the hope of keeping him busy and quiet for a while Gauss immediately came back with the answer 5050 = /, because he saw the following Figure 1 The triangular numbers are t 1 = 1, t = 3, t 3 = 6, t 4 = 10,

3 11 Polygonal Numbers 3 Figure 13 t 1 = 1, t = 3, t 3 = 4 3, t 4 = 5 4, trick, which works for any n Write the sum defining t n twice, once forward and once backward: n 1+n, n+n Now, add vertically; each pair of terms sums to n + 1, and there are n terms, so t n = n(n + 1) or t n = n(n + 1)/ The third proof uses mathematical induction This is a method of proof that works when there are infinitely many theorems to prove, for example, one theorem for each integer n The first case n = 1 must be proven and then it has to be shown that each case follows from the previous one Think about a line of dominoes standing on edge The n = 1 case is analogous to knocking over the first domino The inductive step, showing that case n 1 implies case n, is analogous to each domino knocking over the next one in line We will give a proof of the formula t n = n(n + 1)/ by induction The n = 1 case is easy Figure 1 shows that t 1 = 1, which is equal to (1 )/ Now we get to assume that the theorem is already done in the case of n 1; that is, we can assume that (n 1)n t n 1 = n 1 = So t n = n 1 + n = t n 1 + n = (n 1)n + n = (n 1)n + n (n + 1)n = We have already mentioned the square numbers, s n These are just the number of dots in a square array with n rows and n columns This is easy; the formula is s n = n Nonetheless, the square numbers, s n, are more interesting than one might think For example, it is easy to see that the sum of two consecutive triangular numbers is a square number: Figure 14 shows a geometric proof t n 1 + t n = s n (1)

4 4 1 Sums and Differences Figure 14 Geometric proof of Eq (1) It is also easy to give an algebraic proof of this same fact: t n 1 + t n = (n 1)n + n(n + 1) = (n 1 + n + 1)n = n = s n Figure 11 seems to indicate that we can give an inductive proof of the identity (n 1) = n (13) For the n = 1 case we just have to observe that 1 = 1 And we have to show that the n 1st case implies the nth case But (n 3) + (n 1) = { (n 3)} + n 1 So, by the induction hypothesis, it simplifies to (n 1) + n 1 = n n n 1 = n Exercise 111 Since we know that t n 1 + t n = s n and that (n 1) = s n, it is certainly true that (n 1) = t n 1 + t n Give a geometric proof of this identity That is, find a way of arranging the two triangles for t n 1 and t n so that you see an array of dots in which the rows all have an odd number of dots Exercise 11 Give an algebraic proof of Plutarch s identity 8t n + 1 = s n+1 using the formulas for triangular and square numbers Now give a geometric proof of this same identity by arranging eight copies of the triangle for t n, plus one extra dot, into a square

5 11 Polygonal Numbers 5 Exercise 113 Which triangular numbers are also squares? That is, what conditions on m and n will guarantee that t n = s m? Show that if this happens, then we have (n + 1) 8m = 1, a solution to Pell s equation, which we will study in more detail in Chapter 11 The philosophy of the Pythagoreans had an enormous influence on the development of number theory, so a brief historical diversion is in order Pythagoras of Samos ( BC) Pythagoras traveled widely in Egypt and Babylonia, becoming acquainted with their mathematics Iamblichus of Chalcis, in his On the Pythagorean Life (Iamblichus, 1989), wrote of Pythagoras journey to Egypt: From there he visited all the sanctuaries, making detailed investigations with the utmost zeal The priests and prophets he met responded with admiration and affection, and he learned from them most diligently all that they had to teach He neglected no doctrine valued in his time, no man renowned for understanding, no rite honored in any region, no place where he expected to find some wonderhe spent twenty-two years in the sacred places of Egypt, studying astronomy and geometry and being initiated into all the rites of the gods, until he was captured by the expedition of Cambyses and taken to Babylon There he spent time with the Magi, to their mutual rejoicing, learning what was holy among them, acquiring perfected knowledge of the worship of the gods and reaching the heights of their mathematics and music and other disciplines He spent twelve more years with them, and returned to Samos, aged by now about fifty-six (Cambyses, incidentally, was a Persian emperor who invaded and conquered Egypt in 55 bc, ending the twenty-fifth dynasty According to Herodotus in The Histories, Cambyses did many reprehensible things against Egyptian religion and customs and eventually went mad) The Pythagorean philosophy was that the essence of all things is numbers Aristotle wrote in Metaphysics that [t]hey thought they found in numbers, more than in fire, earth, or water, many resemblances to things which are and become Since, then, all other things seemed in their whole nature to be assimilated to numbers, while numbers seemed to be the first things in the whole of nature, they supposed the elements of numbers to be the elements of all things, and the whole heaven to be a musical scale and a number Musical harmonies, the sides of right triangles, and the orbits of different planets could all be described by ratios This led to mystical speculations about the properties of special numbers In astronomy the Pythagoreans had the concept of the great year If the ratios of the periods of the planets

6 6 1 Sums and Differences Figure 15 The tetrahedral numbers T 1 = 1, T = 4, T 3 = 10, T 4 = 0, are integers, then after a certain number of years (in fact, the least common multiple of the ratios), the planets will return to exactly the same positions again And since astrology says the positions of the planets determine events, according to Eudemus, then Ishall sit here again with this pointer in my hand and tell you such strange things The tetrahedral numbers, T n, are three-dimensional analogs of the triangular numbers, t n They give the number of objects in a tetrahedral pyramid, that is, a pyramid with triangular base, as in Figure 15 The kth layer of the pyramid is a triangle with t k objects in it; so, by definition, T n = t 1 + t + +t n 1 + t n = t k (14) Here, we use Sigma notation to indicate that the kth term in the sum is the kth triangular number, t k What is the pattern in the sequence of the first few tetrahedral numbers: 1, 4, 10, 0,? What is the formula for T n for general n? It is possible to give a three-dimensional geometric proof that T n = n(n + 1)(n + )/6 It helps to use cubes instead of spheres First shift the cubes so they line up one above the other, as we did in two dimensions Then try to visualize six copies of the cubes, which make up T n filling up a box with dimensions n by n + 1 by n + This would be a three- dimensional analog of Figure 13 If this makes your head hurt, we will give another proof that is longer but not so three dimensional In fact you can view the following explanation as a twodimensional analog of Gauss one dimensional proof that t n = n(n + 1)/

7 11 Polygonal Numbers 7 We will do this in the case of n = 5 for concreteness From Eq (14) we want to sum all the numbers in a triangle: The kth row is the triangular number t k We take three copies of the triangle, each one rotated by 10 : The rearranged triangles still have the same sum This is the analog of Gauss taking a second copy of the sum for t n written backward Observe that if we add the left and center triangles together, in each row the sums are constant: = = = = = Inrowk, all the entries are k + 1, just as Gauss found In the third triangle, all the entries in row k are the same; they are equal to n k + 1, and k + 1 plus n k + 1isn = = = = = We get a triangle with t n numbers in it, each of which is equal to n + So, 3T n = t n (n + ) = n(n + 1)(n + )/,

8 8 1 Sums and Differences Figure 16 The pyramidal numbers P 1 = 1, P = 5, P 3 = 14, P 4 = 30, and therefore, T n = n(n + 1)(n + )/6 (15) Exercise 114 Use mathematical induction to give another proof of Eq (15), with T n defined by Eq (14) The pyramidal numbers, P n, give the number of objects in a pyramid with a square base, as in Figure 16 The kth layer of the pyramid is a square with s k = k objects in it; so, by definition, P n = n = k Since we know a relationship between square numbers and triangular numbers, we can get a formula for P n in terms of the formula for T n, as follows From Eq (1) we have t k + t k 1 = k for every k This even works for k = 1 if we define t 0 = 0, which makes sense So, P n = = k = t k + {t k + t k 1 } t k 1 = T n + T n 1 According to Eq (15) this is just P n = n(n + 1)(n + )/6 + (n 1)n(n + 1)/6 = n(n + 1)(n + 1)/6

9 The formulas 11 Polygonal Numbers n = n(n + 1)/, (16) n = n(n + 1)(n + 1)/6 (17) are beautiful Can we generalize them? Is there a formula for sums of cubes? In fact there is, due to Nicomachus of Gerasa Nicomachus observed the interesting pattern in sums of odd numbers: 1 = 1 3, = 3, = 3 3, = 4 3, = 5 3, This seems to indicate that summing consecutive cubes will be the same as summing consecutive odd numbers = , = , But how many odd numbers do we need to take? Notice that 5 is the third odd number, and t = 3 Similarly, 11 is the sixth odd number, and t 3 = 6 We guess that the pattern is that the sum of the first n cubes is the sum of the first t n odd numbers Now Eq (13) applies and this sum is just (t n ) From Eq (11) this is (n(n + 1)/) So it seems as if n 3 = n (n + 1) /4 (18) But the preceding argument was mostly inspired guessing, so a careful proof by induction is a good idea The base case n = 1 is easy because 1 3 = 1 /4 Now we can assume that the n 1 case (n 1) 3 = (n 1) n /4

10 10 1 Sums and Differences Table 11 Another proof of Nicomachus identity is true and use it to prove the next case But (n 1) 3 + n 3 ={ (n 1) 3 }+n 3 = (n 1) n 4 + n 3 by the induction hypothesis Now, put the two terms over the common denominator and simplify to get n (n + 1) /4 Exercise 115 Here s another proof that n 3 = n (n + 1) /4, (19) with the details to be filled in The entries of the multiplication table are shown in Table 11 Each side of the equation can be interpreted as a sum of all the entries in the table For the left side of Eq (19), form gnomons starting from the upper-left corner For example, the second one is, 4, The third one is 3, 6, 9, 6, 3, and so on What seems to be the pattern when you add up the terms in the kth gnomon? To prove your conjecture, consider the following questions: 1 What is the common factor of all the terms in the kth gnomon? If you factor this out, can you write what remains in terms of triangular numbers? 3 Can you write what remains in terms of squares? 4 Combine these ideas to prove the conjecture you made The right side of Eq (19) is t n Why is the sum of the n entries in the first n rows and n columns equal to t n t n?